Legendre Transformations and a decrease in Helmholtz free energy

Question

Suppose we have a system placed in a thermal reservoir, and the system is free to exchange energy but not volume or particles with the reservoir. Under these conditions, it can be shown that an increase in the total entropy of the reservoir + system is equivalent to a decrease in the Helmholtz free energy of the reservoir. In other words, the system will spontaneously tend toward the state of lowest Helmholtz free energy.

This statement, though, seems trivial.

Let $$F(T,...)=-TS+U(S,...)$$ where $T=\partial U / \partial S$ and

To find the minimum value of $F$ differentiate with respect to $T:$

$$\partial F / \partial T = 0=S$$

So this would mean that the reservoir would tend toward the state of zero entropy.

So again, what is meant that a system will spontaneously tend toward the state of lowest Helmholtz free energy?

Answer 1

The minimization of the free energy determines the equilibrium state of a system at fixed $(T,V,N)$. This minimization is done with respect to the partitioning of $V$ and $N$ under fixed temperature. We divide the system into two parts and seek the conditions that minimize the total free energy, $$ d(A_1+A_2) = 0 $$ under the constraints $$ T_1 = T_2 = T,\quad V_1+V_2 = V,\quad N_1+N_2= N. $$ The minimizatio yields: $$ \left(\frac{\partial A_1}{\partial V_1}\right)_{T,N_1} = \left(\frac{\partial A_2}{\partial V_2}\right)_{T,N_1} \Rightarrow \boxed{P_1 = P_2} $$ $$ \left(\frac{\partial A_1}{\partial N_1}\right)_{T,V_1} = \left(\frac{\partial A_2}{\partial N_2}\right)_{T,V_2} \Rightarrow \boxed{\mu_1 = \mu_2} $$ These express the standard conditions of mechanical and chemical equilibrium (thermal equilibrium is maintained by the condition $T_1=T_2$).

Your mistake is to minimize $A$ with respect to $T$. In general, we minimize potentials with respect to their extensive variables while keeping their intensive variables fixed.